Question

in a random sample of six cell phones, the mean full retail price was $579.00 and the standard deviation was $210.00. assume the population is normally distributed and use the t - distribution to find the margin of error and construct a 99% confidence interval for the population mean μ. interpret the results. identify the margin of error. 345.8 dollars (round to one decimal place as needed.) construct a 99% confidence interval for the population mean. (round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The confidence interval for the population mean when using the t - distribution is given by \(\bar{x}\pm E\), where \(\bar{x}\) is the sample mean and \(E\) is the margin of error.

Step2: Substitute the values

We are given that \(\bar{x} = 579\) and \(E=345.8\)

For the lower bound: \(579 - 345.8\)

$$579-345.8=233.2$$

For the upper bound: \(579 + 345.8\)

$$579 + 345.8=924.8$$

Answer:

\((233.2,924.8)\)